Optimal. Leaf size=338 \[ -\frac {x}{4 (a-i b)^{5/3}}-\frac {x}{4 (a+i b)^{5/3}}-\frac {i \sqrt {3} \text {ArcTan}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 (a-i b)^{5/3} d}+\frac {i \sqrt {3} \text {ArcTan}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 (a+i b)^{5/3} d}+\frac {i \log (\cos (c+d x))}{4 (a-i b)^{5/3} d}-\frac {i \log (\cos (c+d x))}{4 (a+i b)^{5/3} d}+\frac {3 i \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a-i b)^{5/3} d}-\frac {3 i \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a+i b)^{5/3} d}-\frac {3 b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{2/3}} \]
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Rubi [A]
time = 0.26, antiderivative size = 338, normalized size of antiderivative = 1.00, number of steps
used = 12, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3564, 3620,
3618, 59, 631, 210, 31} \begin {gather*} -\frac {3 b}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{2/3}}-\frac {i \sqrt {3} \text {ArcTan}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d (a-i b)^{5/3}}+\frac {i \sqrt {3} \text {ArcTan}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d (a+i b)^{5/3}}+\frac {3 i \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d (a-i b)^{5/3}}-\frac {3 i \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d (a+i b)^{5/3}}+\frac {i \log (\cos (c+d x))}{4 d (a-i b)^{5/3}}-\frac {i \log (\cos (c+d x))}{4 d (a+i b)^{5/3}}-\frac {x}{4 (a-i b)^{5/3}}-\frac {x}{4 (a+i b)^{5/3}} \end {gather*}
Antiderivative was successfully verified.
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Rule 31
Rule 59
Rule 210
Rule 631
Rule 3564
Rule 3618
Rule 3620
Rubi steps
\begin {align*} \int \frac {1}{(a+b \tan (c+d x))^{5/3}} \, dx &=-\frac {3 b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{2/3}}+\frac {\int \frac {a-b \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx}{a^2+b^2}\\ &=-\frac {3 b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{2/3}}+\frac {\int \frac {1+i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx}{2 (a-i b)}+\frac {\int \frac {1-i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx}{2 (a+i b)}\\ &=-\frac {3 b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{2/3}}+\frac {\text {Subst}\left (\int \frac {1}{(-1+x) (a+i b x)^{2/3}} \, dx,x,-i \tan (c+d x)\right )}{2 (i a-b) d}-\frac {\text {Subst}\left (\int \frac {1}{(-1+x) (a-i b x)^{2/3}} \, dx,x,i \tan (c+d x)\right )}{2 (i a+b) d}\\ &=-\frac {x}{4 (a-i b)^{5/3}}-\frac {x}{4 (a+i b)^{5/3}}+\frac {i \log (\cos (c+d x))}{4 (a-i b)^{5/3} d}-\frac {i \log (\cos (c+d x))}{4 (a+i b)^{5/3} d}-\frac {3 b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{2/3}}-\frac {(3 i) \text {Subst}\left (\int \frac {1}{\sqrt [3]{a-i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a-i b)^{5/3} d}-\frac {(3 i) \text {Subst}\left (\int \frac {1}{(a-i b)^{2/3}+\sqrt [3]{a-i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a-i b)^{4/3} d}+\frac {(3 i) \text {Subst}\left (\int \frac {1}{\sqrt [3]{a+i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a+i b)^{5/3} d}+\frac {(3 i) \text {Subst}\left (\int \frac {1}{(a+i b)^{2/3}+\sqrt [3]{a+i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a+i b)^{4/3} d}\\ &=-\frac {x}{4 (a-i b)^{5/3}}-\frac {x}{4 (a+i b)^{5/3}}+\frac {i \log (\cos (c+d x))}{4 (a-i b)^{5/3} d}-\frac {i \log (\cos (c+d x))}{4 (a+i b)^{5/3} d}+\frac {3 i \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a-i b)^{5/3} d}-\frac {3 i \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a+i b)^{5/3} d}-\frac {3 b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{2/3}}+\frac {(3 i) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}\right )}{2 (a-i b)^{5/3} d}-\frac {(3 i) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}\right )}{2 (a+i b)^{5/3} d}\\ &=-\frac {x}{4 (a-i b)^{5/3}}-\frac {x}{4 (a+i b)^{5/3}}-\frac {i \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 (a-i b)^{5/3} d}+\frac {i \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 (a+i b)^{5/3} d}+\frac {i \log (\cos (c+d x))}{4 (a-i b)^{5/3} d}-\frac {i \log (\cos (c+d x))}{4 (a+i b)^{5/3} d}+\frac {3 i \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a-i b)^{5/3} d}-\frac {3 i \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a+i b)^{5/3} d}-\frac {3 b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{2/3}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in
optimal.
time = 0.16, size = 106, normalized size = 0.31 \begin {gather*} \frac {3 i \left ((a+i b) \, _2F_1\left (-\frac {2}{3},1;\frac {1}{3};\frac {a+b \tan (c+d x)}{a-i b}\right )-(a-i b) \, _2F_1\left (-\frac {2}{3},1;\frac {1}{3};\frac {a+b \tan (c+d x)}{a+i b}\right )\right )}{4 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{2/3}} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 0.21, size = 101, normalized size = 0.30
method | result | size |
derivativedivides | \(\frac {3 b \left (-\frac {1}{2 \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{\frac {2}{3}}}+\frac {\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (-\textit {\_R}^{3}+2 a \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}}{6 a^{2}+6 b^{2}}\right )}{d}\) | \(101\) |
default | \(\frac {3 b \left (-\frac {1}{2 \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{\frac {2}{3}}}+\frac {\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (-\textit {\_R}^{3}+2 a \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}}{6 a^{2}+6 b^{2}}\right )}{d}\) | \(101\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{3}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 6.44, size = 2500, normalized size = 7.40 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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